The Trigonometric Functions
by
Stefan Waner and Steven R. Costenoble

This Section: 3. Derivatives of Trigonometric Functions

2. The Six Trigonometric Functions

Section 3 Exercises

4. Integrals of Trigonometric Functions

Trigonometric Functions Main Page

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We shall start by giving the derivative of f(x) = sin x, and then using it to obtain the derivatives of the other five trigonometric functions.

Derivative of sin x The derivative of the sine function is given by d dx sin x = cos x.

That's all there is to it!

Question
Where did that come from?

Answer
We shall justify this at the end of this section. (If you can't wait, press the pearl to go there now.)

Example 1

Calculate dy/dx if:

(a) y = x sin x	(b) y = cosec x	(c) y	=	x2+x sin x
(d) y = sin(3x2−1)

Solution

(a) An application of the calculator thought experiment (CTE)^* tells us that x sin x is a product;

y = (x)(sin x).

Therefore, by the product rule,

dy dx

=

(1)(sin x) + (x)(cos x) = sin x + x cos x

(b) Recall from Section 2 that

y = cosec x =

1 sin x

.

Therefore, by the quotient rule,

dy dx	=	(0)(sin x) − (1)(cos x) sin2x	(recall that sin2x is just (sin x)2)
	=	−cos x sin2x
	=	− cos x sin x . 1 sin x
	=	− cotan x cosec x.	(from the identities in Section 2)

Notice that we have just obtained the derivative of one of the remaining five trigonometric functions. Four to go...

(c) Since the given function is a quotient,

dy dx

=

(2x+1)(sin x) − (x2+x)(cos x) sin2x

,

and let us just leave it like that (there is no easy simplification of the answer).

(d) Here, an application of the CTE  tells us that y is the sine of a quantity.

Since

d dx	sin x = cos x,

the chain rule ( press the pearl to go to the topic summary for a quick review) tells us that

d dx

sin u = cos u

du dx

so that

d dx	sin (3x2−1)	=	cos (3x2−1) d dx (3x2−1)
		=	6x cos(3x2−1)		(we placed the 6x in front to avoid confusion—see below)

^* See Example 6 on p. 258 in Calculus Applied to the Real World, or p. 756 in Finite Mathematics and Calculus Applied to the Real World. Alternatively, press here to consult the on-line topic summary, where the CTE is also discussed.

Before we go on...

Try to avoid writing expressions such as cos(3x²−1)(6x). Does this mean

cos[(3x2−1)(6x)]

(the cosine of the quantity (3x2−1)(6x)),

or does it mean

[cos(3x2−1)](6x)

(the product of cos(3x2−1) and 6x)?

That is why we placed the 6x in front of the cosine expression.

Question
What about the derivative of the cosine function?

Answer
Let us use the identity

cos x = sin(π/2−x)

from Section 1, and follow the method of Example 1(d) above: if

y = cos x = sin(π/2−x),

then, using the chain rule,

dy dx	=	cos(π/2−x) d dx (π/2−x)
	=	(−1)cos(π/2−x)	(since π/2 is constant, and d/dx(−x) = −1)
	=	−sin x	(using the identity cos(π/2−x) = sin x)

Question
And the remaining three trigonometric functions?

Answer
Since all the remaining ones are expressible in terms of sin x and cos x, we shall leave them for you to do in the exercises!

Original Rule	Generalized Rule (Chain Rule)
d dx sin x = cos x	d dx sin u = cos u du dx
d dx cos x = −sin x	d dx cos u = −sin u du dx
d dx tan x = sec2 x	d dx tan u = sec2u du dx
d dx cotan x = −cosec2 x	d dx cotan u = −cosec2u du dx
d dx sec x = sec x tan x	d dx sec u = sec u tan u du dx
d dx cosec x = −cosec x cotan x	d dx cosec u = −cosec u cotan u du dx

Example 2

Find the derivatives of the following functions.

(a) f(x) = tan(x2−1)	(b) g(x) = cosec(e3x)	(c) h(x) = e−xsin(2x)
(d) r(x) = sin2x	(e) s(x) = sin(x2)

Solution

(a) Since f(x) is the tan of a quantity, we use the chain rule form of the derivative of tangent:

d dx

tan u = sec2u

du dx

d dx	tan(x2−1)	=	sec2(x2−1)	d(x2−1) dx	(substituting u = x2−1)
		=	2x sec2(x2−1).

(b) Since g(x) is the cosecant of a quantity, we use the rule

d dx

cosec u = −cosec u cotan u

du dx

d dx	cosec(e3x)	=	−cosec(e3x) cotan(e3x)	d(e3x) dx
		=	−3e3x cosec(e3x) cotan(e3x).		(the derivative of e3x is 3e3x)

(c) Since h(x) is the product of e^−x and sin(2x), we use the product rule,

h'(x)	=	(−e−x)sin(2x) + e−x d dx [sin(2x)]
	=	(−e−x)sin(2x) + e−x cos(2x) d dx [2x]	(using d/dx sin u = cos u du/dx)
	=	−e−xsin(2x) + 2e−xcos(2x).

(d) Recall that sin²x = (sin x)². Thus, r(x) is the square of a quantity (namely, the quantity sin x). Therefore, we use the chain rule for differentiating the square of a quantity,

d dx

[u2]

=

2u

du dx

d dx	[(sin x)2]	=	2(sin x)	d(sin x) dx
		=	2 sin x cos x.

(e) Notice the difference between sin²x and sin(x²). The first is the square of sin x, while the second is the sin of the quantity x². Since we are differentiating the latter, we use the chain rule for differentiating the sine of a quantity:

d dx

sin u = cos u

du dx

d dx	sin(x2)	=	cos(x2)	d(x2) dx
		=	2x cos(x2).

and we are left with two limits to evaluate. Calculating these limits analytically requires a little trigonometry (press here for these calculations). Alternatively, we can get a good idea of what these two limits are by estimating them numerically. We find that:

h0

(cos h − 1) h

=

0

and

h0

sin h h

=

1

Therefore,

d dx	sin x = sin x (0) + cos x (1) = cos x.

2. The Six Trigonometric Functions

Section 3 Exercises

4. Integrals of Trigonometric Functions

Trigonometric Functions Main Page

"RealWorld" Page

Everything for Calculus